Question

\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]

Answer 1

\[\int\frac{\cos \left( 2 x \right) \cdot dx}{\sqrt{\sin^2 2x + 8}}\]
\[\text{ let } \text{ sin } \left( 2x \right) = t\]
\[ \Rightarrow \text{ cos }\left( 2x \right) \times 2 \cdot dx = dt\]
\[ \Rightarrow \text{ cos }\left( 2x \right) \cdot dx = \frac{dt}{2}\]
\[Now, \int\frac{\text{ cos } \left( 2 x \right) \cdot dx}{\sqrt{\sin^2 2x + 8}} \]
\[ = \frac{1}{2}\int\frac{dt}{\sqrt{t^2 + \left( 2\sqrt{2} \right)^2}}\]
\[ = \frac{1}{2}\text{ log }\left| t + \sqrt{t^2 + 8} \right| + C\]
\[ = \frac{1}{2} \text{ log }\left| \text{ sin }\left( 2x \right) + \sqrt{\text{ sin }^2 \left(\text{  2x }\right) + 8} \right| + C\]